Measuring Space: Perimeter and Area: questions with answers

The wire never changed length, so both shapes have the same perimeter: the square gives $4 \times 10 = 40$ cm and the rectangle gives $2(8 + 12) = 40$ cm. Their areas are not the same. The square holds $10^2 = 100$ square centimetres and the rectangle holds $8 \times 12 = 96$, which is 4 less. A fixed length of wire does not fix the area it can enclose.

Both rectangles have perimeter 40 cm, so the border tells you nothing about the inside. Rectangle P covers $2 \times 18 = 36$ square centimetres and rectangle Q covers $8 \times 12 = 96$, so Q exceeds P by 60 square centimetres. Equal perimeter does not mean equal area: the nearer a rectangle of fixed perimeter comes to a square, the more it holds.

Every circle has the same circumference-to-diameter ratio, so the second circumference is $44 \div 14 \times 70$, which is just 5 times the first. The diameter went from 14 cm to 70 cm, a factor of 5, and the circumference follows it exactly. It does not grow by $5^2 = 25$; that would mean bigger circles had a bigger C to D ratio, and they do not.

The circumference is $\pi$ times the DIAMETER: $\frac{22}{7} \times 21 = 66$ cm. The other form, $2\pi r$, uses the radius, and the two are the same rule because the diameter is twice the radius. Putting the diameter into the radius form gives 132, which is twice too long.

The repeating block is $45$, and it has two digits. A repeating block is written over as many $9$s as it has digits, so the fraction is $\frac{45}{99}$, which is $\frac{5}{11}$. Dividing $5$ by $11$ does give $0.454545\ldots$, so the decimal is rational even though its digits never stop. Compare $\frac{1}{7} = 0.142857\,142857\ldots$, which also never stops and is plainly a fraction. Only a decimal that never falls into a pattern, like $\pi$, has no fraction equal to it.

No fraction of whole numbers is exactly equal to $\pi$, and that is what calling $\pi$ irrational means. $\frac{22}{7}$ is about $3.142857$, so it parts from $\pi = 3.14159...$ at the third decimal place. $\frac{355}{113}$ is about $3.1415929$, which is closer, but it parts from $\pi$ at the seventh decimal place. And $3.1416$ is only the fraction $\frac{31416}{10000}$, a rounded value. For any fraction close to $\pi$ we can always find a closer one, so there is no best fraction and no exact one.

Each arc is its own fraction of the same circumference, so the extra length is the extra fraction of it: the second arc runs 180 degrees further round, which is $\frac{180}{360}$ of the circumference 704 cm, or 352 cm. Taking that fraction of the RADIUS instead gives 56, which slices the wrong quantity: an arc is a piece of the way ROUND the circle, not of the way out to its edge.

Of the 564 cm walked round the sector, the two straight edges account for $2 \times 84 = 168$ cm. Taking them off leaves the arc: $564 - 168 = 396$ cm. Answering 564 would mean the whole border was curved, and a sector always has two straight edges as well.

The rectangle covers $4 \times 9 = 36$ square centimetres and the square covers $6 \times 6 = 36$ square centimetres, so the area is the same for both. Nothing else matches. The shapes are plainly different, and the perimeters differ too: $2(4 + 9) = 26$ cm against $4 \times 6 = 24$ cm. The sides differ as well, $4$ cm and $9$ cm against $6$ cm. Area counts how much a region covers; it does not fix the outline. Baudhayana's construction in this chapter turns a rectangle into a square of exactly the same area, which is this same point made with a compass.

The first rectangle has perimeter $2(6 + 6) = 24$ cm and the second has perimeter $2(4 + 9) = 26$ cm. Both cover $36$ square centimetres, yet their perimeters differ. A $1$ cm by $36$ cm strip covers $36$ too, with a perimeter of $74$ cm. Fixing the area leaves the shape free, and the perimeter moves with the shape.

Both triangles stand on the base $BC$. Both apexes lie on a line parallel to $BC$, so each apex is the same perpendicular distance from the base. Same base and same height, so half base times height gives the same answer for both. Being more slanted changes the shape of $\triangle QBC$, not its height, so the two areas are equal.

The height is the perpendicular distance from the apex to the base, and that is 7 cm whether or not its foot lands on the segment itself. So the area is $\frac{1}{2} \times 12 \times 7 = 42$ square centimetres. Using the side $AB = 10$ cm as the height gives 60, which is too large: the perpendicular is the shortest distance from $A$ to the line, so any side reaching $A$ is longer than it.

The diagonals bisect each other at right angles, forming four right triangles with perpendicular sides $3$ cm and $4$ cm. Each triangle has area $\frac{1}{2} \times 3 \times 4 = 6$ square centimetres. The rhombus therefore has area $4 \times 6 = 24$ square centimetres. A square with the same four side lengths has area $25$ square centimetres. Equal side lengths alone do not fix a quadrilateral's area.

Brahmagupta’s formula requires a cyclic quadrilateral: all four vertices lie on one circle. Four side lengths alone do not generally determine a quadrilateral’s area. The cyclic condition supplies the needed restriction.

The formula squares the RADIUS, so halve first: the radius is $28 \div 2 = 14$ cm and the area is $\frac{22}{7} \times 14^2 = 616$ square centimetres. Squaring the diameter instead gives 2464, which is four times too large, because the mistaken factor of two is squared along with everything else.